Micromass' big August challenge

[micromass]

I didn't find a proof that every sequence has to lead to convergence after a finite number of steps.

Right, and you can't find such a proof. It's a good idea though, but it will fail to provide a generalized limit on every sequence. Note that generalized limits are not unique either!

 

[mfb]

If there is no such proof (and if you say that I trust you), then there should be a counterexample, but I don't find one.

 

[micromass]

If there is no such proof (and if you say that I trust you), then there should be a counterexample, but I don't find one.

It might actually be so that AC might need to be invoked to find a counterexample. I know for sure your statement is false with AC though, but I will not show it until this question is solved.

 

[micromass]

This doesn't spoil anything, but it's a counterexample to the claim made by mfb: http://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-cesàro-mean

 

[mfb]

This doesn't spoil anything, but it's a counterexample to the claim made by mfb: http://mathoverflow.net/questions/84772/bounded-sequences-with-divergent-cesàro-mean

Thanks. So the point is to blow up the length of the sub-sequences fast enough. Probably faster than e^e^x or something like that (didn't check it in detail).

 

[fresh_42]

Thanks. So the point is to blow up the length of the sub-sequences fast enough. Probably faster than e^e^x or something like that (didn't check it in detail).

This sounds like transforming the issue until you end up transcending something. (At least this would give us a reason for AC.)
I was thinking about another idea: All difficulties arise either from linearity or (if this is solved) from $\underline{\lim} \leq L \leq \overline{\lim}$.
How about putting AC in a Hamel basis (if needed at all) and define $L(x_n)$ as a certain derivation? (Don't know whether this is a real idea or simply a crude thought.)

 

[mfb]

Hamel basis is a good keyword.

We can solve parts of the linearity by directly using this linearity in the definition. Find a generalized limit in [-1,1] for all sequences where $\liminf_n x_n = -1$ and $\limsup_n x_n = 1$. Then scale sequence and generalized limit linearly to find the generalized limit of sequences which have other liminf and limsup. This doesn't work for convergent series, where we set the generalized limit to the regular limit separately.

Now the interesting question: does this procedure survive condition (1), the limit of the sum of two sequences? I guess this depends on the definition. The critical point here is a cancellation of accumulation points, e.g. (0,2,-1,2,0,2,-1,...) + (0,-1,0,0,0,-1,0,0,0,-1,...) and (0,2,-1,2,0,2,-1,...) + (0,-1,0,0,0,-1,0,0,0,-1,...) have different limsup but need the same limit (condition 5 plus condition 1).

 

[fresh_42]

I'll have a try on #4, although I admit that my solution hides AC in a general result, which I won't prove here. And my first idea hasn't been that far from what I have (not exactly a derivation, but something linear instead). The whole problem is to connect the algebraic property of linearity with the topological property of the $\lim \inf$ and $\lim \sup$.

I'll start with a description of Hewitt, Stromberg, Real and Abstract Analysis, Springer in which the reals are defined as the cosets of a Cauchy sequence $\mathfrak{C}$ modulo null sequences $\mathfrak{N}$. Both are subsets of all bounded real sequences $\mathfrak{B}$.

With $\mathfrak{N} \subset \mathfrak{C} \subset \mathfrak{B}$ we get $\mathbb{R} \cong \mathfrak{C} / \mathfrak{N} \subset \mathfrak{B} / \mathfrak{N} =: \mathfrak{A}$ a real (topological, infinite dimensional, commutative) $\mathbb{R}-$algebra $\mathfrak{A}$ (with $1$).

The linear function $L$ can now be viewed as a linear functional on $\mathfrak{A}$.
I will use the sub-additivity $\overline{\lim}((x_n)+(y_n)) \leq \overline{\lim} (x_n)+\overline{\lim} (y_n)$ and $\underline{\lim}(x_n) = - \overline{\lim}(-x_n)$

Now here's the trick and the point where the real work is done:
If we define $L(x_n)\vert_{\mathfrak{C} / \mathfrak{N}} := \lim(x_n)$ then $L(x_n) = \lim(x_n) \leq \overline{\lim}(x_n)$ holds on $\mathfrak{C} / \mathfrak{N} \subset \mathfrak{A} \;$, is linear and we may apply the theorem of Hahn-Banach. (Its proof uses Zorn's Lemma and therefore AC.)
This theorem guarantees us the existence of a linear functional $L$ on $\mathfrak{A}$ which extends our $L$ defined on $\mathfrak{C} / \mathfrak{N} \;$.

Linearity is already given by Hahn-Banach.
With that we have $L(x_n) \leq \overline{\lim}(x_n)$ by the theorem and
$$L(x_n)=-L(-x_n) \geq -\overline{\lim}(-x_n) = \underline{\lim}(x_n)$$
For the cutting process in point 5) we define a sequence $z_n$ by $z_1=x_1$ and $z_n=x_{n} - y_{n-1}=0$ for $n > 1$.
Then $L(z_n)=L(x_1,0, \dots) = 0 = L(x_n) - L(0,y_1,y_2, \dots) = L(x_n) - L(y_n)$ and 5) is equivalent to $L(0,y_1,y_2, \dots) = L(y_1,y_2, \dots)$. This is true for elements of $\mathfrak{C} / \mathfrak{N}$ which are dense in $\mathfrak{A}$ according to $\Vert \, . \Vert_{\infty}$.

For $0 \leq x_n$ we have $0 \leq \underline{\lim} (x_n) \leq L(x_n)$; and for $\lim (x_n) = x$ we have $x = \underline{\lim} (x_n) \leq L(x_n) \leq \overline{\lim} (x_n) = x$ and get equality everywhere. Therefore 4) and 6) hold.

 

[Erland]

Nice fresh42, I should have thought of Hahn-Banach...

 

[micromass]

Here is the solution for #2 of the previously unsolved advanced challenges. I got this from the excellent probability book from Feller, which contains a lot of other gems.

Take the random distributions of $r$ balls in $n$ cells assuming that each arrangement has probability $1/n^r$. We seek the probability$p_m(r,n)$ that exactly $m$ cells are empty.

Let $A_k$ be the event that cell number $k$ is empty. In this even all $r$ balls are placed in the remaining $n-1$ cells and this can be done in $(n-1)^r$ different ways. Similarly, there are $(n-2)^r$ arrangments leaving two preassigned cells empty. By the inclusion-exclusing rule we get that the probability that all cells are occupied is
$$p_0(r,n) = \sum_{\nu = 0}^n (-1)^\nu \binom{n}{\nu} \left(1-\frac{\nu}{n}\right)^r$$
Consider a distribution in which exactly $m$ cells are empty. These $m$ cells can be chosen in $\binom{n}{m}$ ways. The $r$ balls are distributed among the remaining $n-m$ cells so that each of these cells is occupied. The number of such distributions is $(n-m)^r p_0(r,n-m)$. Dividing by $n^r$ we find for the probability that exactly $m$ cells remain empty
$$p_m(r,n)=\binom{n}{m}\sum_{\nu=0}^{n-m} (-1)^\nu \binom{n-m}{\nu} \left(1-\frac{m+\nu}{n}\right)^r$$

Now we wish to find an asymptotic distribution. First note that
$$(n-\nu)^r<n(n-1)...(n-\nu+1)<n^r$$
Thus we have
$$n^\nu\left(1 - \frac{\nu}{n}\right)^{\nu+r}<\nu!\binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r < n^\nu\left(1-\frac{\nu}{n}\right)^r$$
For $0<t<1$, it is clear that $t<-\log(1-t)<\frac{t}{1-t}$ and thus that
$$\left(n e^{-(\nu+r)/(n-\nu)}\right)^\nu <\nu!\binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r < \left(ne^{-r/n}\right)^\nu$$
Now we put $\lambda = ne^{-r/n}$ and we assume that $r$ and $n$ increase in such a way that $\lambda$ remained constrained to a finite interval $0<a<\lambda<b$. For each fixed $\nu$ the ratio of the extreme members in the inequality tends to unity and so
$$0\leq \frac{\lambda^\nu}{\nu!} - \binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r\rightarrow 0$$
This relation holds trivially when $\lambda\rightarrow 0$ and hence it remains true whenever $r$ and $n$ increase such a way that $\lambda$ remains bounded. Now
$$e^{-\lambda} - p_0(r,n) =\sum_{\nu=0}^{+\infty} (-1)^\nu\left(\frac{\lambda^\nu}{\nu!} - \binom{n}{\nu}\left(1-\frac{\nu}{n}\right)^r\right)$$
and the right hand side tends to zero. We therefore have that for each $m$ fixed
$$p_m(r,n) - e^{-\lambda}\frac{\lambda^m}{m!}\rightarrow 0$$

 

[Erland]

There seems to be an error in the given proof of Advanced Problem 4:

I'll have a try on #4, although I admit that my solution hides AC in a general result, which I won't prove here. And my first idea hasn't been that far from what I have (not exactly a derivation, but something linear instead). The whole problem is to connect the algebraic property of linearity with the topological property of the $\lim \inf$ and $\lim \sup$.

I'll start with a description of Hewitt, Stromberg, Real and Abstract Analysis, Springer in which the reals are defined as the cosets of a Cauchy sequence $\mathfrak{C}$ modulo null sequences $\mathfrak{N}$. Both are subsets of all bounded real sequences $\mathfrak{B}$.

With $\mathfrak{N} \subset \mathfrak{C} \subset \mathfrak{B}$ we get $\mathbb{R} \cong \mathfrak{C} / \mathfrak{N} \subset \mathfrak{B} / \mathfrak{N} =: \mathfrak{A}$ a real (topological, infinite dimensional, commutative) $\mathbb{R}-$algebra $\mathfrak{A}$ (with $1$).

The linear function $L$ can now be viewed as a linear functional on $\mathfrak{A}$.
I will use the sub-additivity $\overline{\lim}((x_n)+(y_n)) \leq \overline{\lim} (x_n)+\overline{\lim} (y_n)$ and $\underline{\lim}(x_n) = - \overline{\lim}(-x_n)$

Now here's the trick and the point where the real work is done:
If we define $L(x_n)\vert_{\mathfrak{C} / \mathfrak{N}} := \lim(x_n)$ then $L(x_n) = \lim(x_n) \leq \overline{\lim}(x_n)$ holds on $\mathfrak{C} / \mathfrak{N} \subset \mathfrak{A} \;$, is linear and we may apply the theorem of Hahn-Banach. (Its proof uses Zorn's Lemma and therefore AC.)
This theorem guarantees us the existence of a linear functional $L$ on $\mathfrak{A}$ which extends our $L$ defined on $\mathfrak{C} / \mathfrak{N} \;$.

Linearity is already given by Hahn-Banach.
With that we have $L(x_n) \leq \overline{\lim}(x_n)$ by the theorem and
$$L(x_n)=-L(-x_n) \geq -\overline{\lim}(-x_n) = \underline{\lim}(x_n)$$
For the cutting process in point 5) we define a sequence $z_n$ by $z_1=x_1$ and $z_n=x_{n} - y_{n-1}=0$ for $n > 1$.
Then $L(z_n)=L(x_1,0, \dots) = 0 = L(x_n) - L(0,y_1,y_2, \dots) = L(x_n) - L(y_n)$ and 5) is equivalent to $L(0,y_1,y_2, \dots) = L(y_1,y_2, \dots)$. This is true for elements of $\mathfrak{C} / \mathfrak{N}$ which are dense in $\mathfrak{A}$ according to $\Vert \, . \Vert_{\infty}$.

For $0 \leq x_n$ we have $0 \leq \underline{\lim} (x_n) \leq L(x_n)$; and for $\lim (x_n) = x$ we have $x = \underline{\lim} (x_n) \leq L(x_n) \leq \overline{\lim} (x_n) = x$ and get equality everywhere. Therefore 4) and 6) hold.

The problem lies in this line:

$L(0,y_1,y_2, \dots) = L(y_1,y_2, \dots)$. This is true for elements of $\mathfrak{C} / \mathfrak{N}$ which are dense in $\mathfrak{A}$ according to $\Vert \, . \Vert_{\infty}$.

It is not clear what you mean by $\Vert \, . \Vert_{\infty}$. Normally, it would mean that $\Vert \, (x_n) _n\Vert_{\infty}=\sup_n|x_n|$, but to make sense here, I suppose it means $\Vert \, (x_n)_n\Vert_{\infty}=\overline\lim |x_n|$.
Anyway, it is certainly wrong that $\mathfrak{C} / \mathfrak{N}$ is dense in $\mathfrak{A}$ with respect to this. For example, let $x_n=(-1)^n$ ($n=1,2,\dots$). Then, there is no Cauchy sequence $(y_n)_n$ in $\mathbb R$ such that $\Vert (y_n-x_n)_n\Vert_\infty< 1$, for both these possible definitions of $\Vert \, . \Vert_{\infty}$.
Thus, $\mathfrak{C} / \mathfrak{N}$ is not dense in $\mathfrak{A}$.

So, the proof that 5) holds for the extended functional fails. And in fact, there is no obvious reason that an extended functional, which is proved to exist by Hahn-Banach's Theorem, should satisfy 5) just because the original (unextended) functional does.

Advanced Problem 4 must therefore still be considered as open (at least for us) until this is fixed.

 

[micromass]

@fresh_42 do you have an answer to this? If not I will put it back as an open problem.

 

[fresh_42]

How about an algebraic argument?

Let $S$ denote the linear operator that inserts a zero in the first position, i.e. $S(y_1,y_2, \dots ) = S(0,y_1,y_2, \dots )$

Then $L_0 : S.(\mathfrak{C}/\mathfrak{N}) \rightarrow \mathbb{R}$ with $L_0(0,(x_n)) := L((0,(x_n))$ is a linear functional on $\{(0,x_n) \vert (x_n) \in \mathfrak{C}/\mathfrak{N}\}$ which is identical to $L$ on Cauchy sequences ($L$ as defined in the solution above) since $L(0,x_n) = L(x_n)$ for Cauchy sequences $(x_n)$.

Especially $L_0$ is majorized by $L$, i.e. $L_0(0,(x_n)) \leq L(x_n)$. Thus $L_0$ can be linearly extended on $S.\mathfrak{A}$ by Hahn-Banach. (The fact, that the first component is zero, makes all sets linear subspaces.)
Thus $L_0(0,(y_n)) \leq L(y_n)$ for all $(0,(y_n)) \in S.\mathfrak{A}$. But this can only be true if it is an equation, because we can take $(-y_n)$ and linearity to turn it into $L_0(0,(y_n)) \geq L(y_n)$. Hence $L_0(0,(y_n)) = L(y_n)$ for all $(y_n) \in \mathfrak{A}$ and $L_0$ can be expanded by $L_0((y_n)) := L(y_n)$. The new $L_0$ then automatically fulfills all properties.

Remark: I hope there is no circular conclusion in it. And I assume there is a topological proof, too, as $S$ is linear, has norm $1$ and is thus uniformly continuous. One could probably even take a $\mathfrak{C}/\mathfrak{N}-$ Hamel basis of $\mathfrak{A}$ to extend $LS$, because $1 \in \mathfrak{A}$ can be chosen as first basis vector. The important property of Hahn-Banach is the majorizing sublinear function.

 

[Erland]

Ok, it's late night in Sweden and I'm tired, but I still can't see that you solved the problem. Your notation is a little confusing, but as far as I can see, you actually did nothing but defined $L_0$ on $\mathfrak A$ by $L_0(y_1,y_2,y_3,...)=L(y_2,y_3,y_4,...)$, and I don't see that $L_0$ satisfies 5).

Please, correct me if I am wrong...

 

[fresh_42]

Ok, it's late night in Sweden and I'm tired, but I still can't see that you solved the problem. Your notation is a little confusing, but as far as I can see, you actually did nothing but defined $L_0$ on $\mathfrak A$ by $L_0(y_1,y_2,y_3,...)=L(y_2,y_3,y_4,...)$, and I don't see that $L_0$ satisfies 5).

Please, correct me if I am wrong...

I'm afraid we share the same time zone ...

The argument goes as follows:
(I only have to consider the case $L(0,y_1,y_2, \dots )= L(y_1,y_2, \dots)$ due to linearity (see my first post on this). And this equation is already true for Cauchy sequences.)

I take the function $L$ as previously defined, i.e. it has all properties except (5).
This means it is linear and defined on all bounded sequences (modulo the ideal of null sequences) which I denote as $\mathfrak{A}$.
$\mathbb{R} \cong \mathfrak{C}/\mathfrak{N}$ are the Cauchy sequences (modulo null sequences), which are a subspace of $\mathfrak{A}$.

Now $\{0\} \times \mathfrak{A}$ defines a vector space with $\{0\} \times \mathfrak{C}/\mathfrak{N}$ as a subspace.
$L_0(0,(x_n)) := L(0,(x_n)) = L((x_n)) \leq L((x_n))$ defines a linear functional on this subspace.

I now apply the theorem of Hahn-Banach (without property (5)) to $L_0$ which is majorized by the linear and therewith sublinear functional $L$ on $\{0\} \times \mathfrak{C}/\mathfrak{N}$ to get a linear extension of $L_0$ on $\{0\} \times \mathfrak{A}$.

Thus $L_0(0,(y_n)) \leq L((y_n))$ for all $(0,y_n) \in \{0\} \times \mathfrak{A}$.
Therefore $L_0(0,-(y_n)) \leq L(-(y_n))$ and $L_0(0,(y_n)) \geq L((y_n))$ because $L_0$ and $L$ are linear.
I then have $L_0(0,(y_n)) = L((y_n))$ on all $(0,y_n) \in \{0\} \times \mathfrak{A}$ which means on all $(y_n) \in \mathfrak{A}$.

Now applying Hahn-Banach again to $\{0\} \times \mathfrak{A} \subset \mathfrak{A}$ and $L_0(0,(y_n)) = L((y_n))$ gives with the same argument / trick as above a linear functional $L'_0$ on $\mathfrak{A}$ with $L'_0(0,(y_n)) = L_0(0,(y_n)) = L((y_n))$ and $L'_0((y_n)) = L((y_n))$.

(Maybe the last argument can stand alone without the previous $L_0$ but as you said, it's too late.)

 

[Erland]

How do you get $L'_0((y_n))=L((y_n))$?

 

[fresh_42]

How do you get $L'_0((y_n))=L((y_n))$?

The same way as before.
$L_0(0,(y_n)) = L((y_n) \leq L((y_n))$ on $\{0\} \times \mathfrak{A}$.
Hahn-Banach gives $L'_0((y_n)) \leq L((y_n))$ on $\mathfrak{A}$.
Now $L'_0(-(y_n)) \leq L(-(y_n))$ and by linearity $L'_0((y_n)) \geq L((y_n))$, i.e. $L'_0((y_n)) = L((y_n))$ on $\mathfrak{A}$.
Since $L'_0$ extends $L$ (5) still holds.

Edit: But you have been absolutely right. I tried to get to $L'_0$ in one step and had trouble with an egg-chicken kind of problem in defining it (and which I didn't even realize in the first place). The trick with the subspace $\{0\} \times \mathfrak{A} \subset \mathfrak{A}$ seems to be an easy way out of this circle. In addition I have to thank you for the insights into the power of this theorem. I've found quite a bit different versions and applications on the way. And I now know that Banach is the keyword of first choice whenever topology meets linearity.

 

[Erland]

Perhaps I am slow-witted, but I just don't get it.

The same way as before.
$L_0(0,(y_n)) = L((y_n) \leq L((y_n))$ on $\{0\} \times \mathfrak{A}$.
Hahn-Banach gives $L'_0((y_n)) \leq L((y_n))$ on $\mathfrak{A}$.

How can you go from $L_0(0,(y_n)) = L((y_n) \leq L((y_n))$ to $L'_0((y_n)) \leq L((y_n))$? I suppose you use that $L'_0(0,(y_n))=L_0(0,(y_n))$, but then, you don't have the same upper estimation functional before and after you apply Hahn-Banach (unless $L$ already satisfies 5). To get an upper estimate by $L((y_n))$ after Hahn-Banach, you should have $L(0,(y_n))$ before Hahn-Banach, while in fact you have $L((y_n))$, which is not the same if $L$ does not satisfy 5).

 

[fresh_42]

I'm not sure I understand you.
First I construct an $L$ on $\mathfrak{C}/\mathfrak{N}$ which satisfies all properties, (5) including.
Then I expand $L$ on $\mathfrak{A}$ say $L_1$.
$L_1$ satisfies all properties except (5).
Now I take $L_1$ on $\{0\} \times \mathfrak{C}/\mathfrak{N} \subset \{0\} \times \mathfrak{A}$ with $L_1 (0,(y_n)) \leq L(y_n)$.
Then I expand $L_1$ on $\{0\} \times \mathfrak{A}$, say $L_2$ which still satisfies $L_2(0,(y_n)) \leq L_1(0,(y_n)) = L(y_n)$.
And now I expand $L_2$ from $\{0\} \times \mathfrak{A}$ on $\mathfrak{A}$, say $L_3$.
It still satisfies $L_3(y_n) \leq L(y_n)$ and its restriction on $\{0\} \times \mathfrak{A}$ is $L_2$

At every step linearity guarantees that "$\leq$" is actually an equation and $L_3(y_n) = L(y_n) = L_2(0,(y_n)) = L_3(0,(y_n))$ by construction.

 

[Erland]

I'm not sure I understand you.
First I construct an $L$ on $\mathfrak{C}/\mathfrak{N}$ which satisfies all properties, (5) including.
Then I expand $L$ on $\mathfrak{A}$ say $L_1$.
$L_1$ satisfies all properties except (5).
Now I take $L_1$ on $\{0\} \times \mathfrak{C}/\mathfrak{N} \subset \{0\} \times \mathfrak{A}$ with $L_1 (0,(y_n)) \leq L(y_n)$.
Then I expand $L_1$ on $\{0\} \times \mathfrak{A}$, say $L_2$ which still satisfies $L_2(0,(y_n)) \leq L_1(0,(y_n)) = L(y_n)$.
And now I expand $L_2$ from $\{0\} \times \mathfrak{A}$ on $\mathfrak{A}$, say $L_3$.
It still satisfies $L_3(y_n) \leq L(y_n)$ and its restriction on $\{0\} \times \mathfrak{A}$ is $L_2$

At every step linearity guarantees that "$\leq$" is actually an equation and $L_3(y_n) = L(y_n) = L_2(0,(y_n)) = L_3(0,(y_n))$ by construction.

You write "It still satisfies $L_3(y_n) \leq L(y_n)$", meaning that it did that before. But, as far as I can see, $L_2$ does not need to satisfy $L_2(y_n) \leq L(y_n)$ for $(y_n)\in \{0\} \times \mathfrak{A}$. What it satisfies is $L_2(0,(y_n)) \leq L(y_n)$.

A correct application of Hahn-Banach would lead from $L_2(0,(y_n)) \leq L(y_n)$ to something like $L_3(x,(y_n))\le L(y_n)$, for $x\in \mathbb R$, as I understand it.

Btw, I have no problem with your clever trick of making an equality into an equation. It is the original inequality I doubt.

 

[fresh_42]

$L_2$ isn't even defined on $\mathfrak{A}$ and you are right. I only have $L_2(0,y_n) = L(y_n)$.
But $L_3$ is defined on all $\mathfrak{A}$ with $L_2$ being its restriction on the subspace $\{0\} \times \mathfrak{A}$.
$L$ always serves as the (sub)linear majorizing function, and linearity to get rid of the $\leq$.
So in the end $L_3(0,y_n) = L_2(0,y_n) = L(y_n) = L_3(y_n)$.

It is the original inequality I doubt.

Why? For Cauchy sequences we have convergence and $L$ is the usual limit which doesn't change, if we put a zero at the start.

 

[Erland]

So in the end $L_3(0,y_n) = L_2(0,y_n) = L(y_n) = L_3(y_n)$.

I don't understand why $L_3(y_n)$ should be equal to the other three expressions.

 

[fresh_42]

I don't understand why $L_3(y_n)$ should be equal to the other three expressions.

Because $L_3$ is constructed as extension of $L_2 \leq L$ from $(0,\mathfrak{A})$ on $\mathfrak{A}$.
The restriction makes it equal to $L_2$ which is still equal to $L$ and the expansion makes it equal to $L$. Equality is actually never lost in the process.

Maybe I've overlooked a logical pit, although I can't see one and I admit it looks a bit like cheating.
I'm still convinced that there might be an easier way to show it. Perhaps with linearity and continuity of the shift-operator and a limit procedure.
Also Hewitt, Stromberg remark on their proof of Hahn-Banach that it could be proven with linear extensions of a Hamel basis to the cost of the upper bounding sublinear function. The latter requires the existence of a minimal element by Zorn's Lemma. Since we have equality in our case, a simple linear extension could be enough. But I'm not very used to Hamel basis so I might not see the traps there.

 

[Erland]

Because $L_3$ is constructed as extension of $L_2 \leq L$ from $(0,\mathfrak{A})$ on $\mathfrak{A}$.
The restriction makes it equal to $L_2$ which is still equal to $L$ and the expansion makes it equal to $L$. Equality is actually never lost in the process.

But you didn't have equality in the first place! $L_2$ is not equal to $L$ (restricted to $\{0\}\times \mathfrak A$). You have $L_2(0,(y_n))=L(y_n)$, not $L_2(y_n)=L(y_n)$ and not $L_2(0,(y_n))=L(0,(y_n))$ (at least I don't see you proved it), because $(y_n)$ and $(0,(y_n))$ is not the same thing.

 

[fresh_42]

But you didn't have equality in the first place! $L_2$ is not equal to $L$ (restricted to $\{0\}\times \mathfrak A$). You have $L_2(0,(y_n))=L(y_n)$, not $L_2(y_n)=L(y_n)$ and not $L_2(0,(y_n))=L(0,(y_n))$ (at least I don't see you proved it), because $(y_n)$ and $(0,(y_n))$ is not the same thing.

That is right. But I don't see where I would need it.

I have
a) $L(0,y) = L(y)$ on $C/N \quad$ [property of limits]
b) $L_1(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_1 \vert_{C/N} = L \leq L$ for $C/N \subset A$]+[linearity]
c) $L_1(0,y) = L(0,y) = L(y)$ on $(0,C/N) \subset A \quad$ [first equation in $A$, second in $C/N$]
d) $L_2(0,y) = L(y)$ on $(0,A) \quad$ [Hahn-Banach with $L_2 \vert_{(0,C/N)} = L_1 \leq L$ for $(0,C/N) \subset (0,A)$]+[linearity]
e) $L_3(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_3 \vert_{(0,A)} = L_2 \leq L$ for $(0,A) \subset A$]+[linearity]

Now $L_3(y) = L(y)$ on $A$ and $L_3(0,y) = L_3 \vert_{(0,A)} (0,y) = L_2 (0,y) = L(y)$.
Since both are equal to $L(y) \;$, $\; L_3(0,y) = L_3(y)$ for all $y \in A$.

 

[Erland]

That is right. But I don't see where I would need it.

I have
a) $L(0,y) = L(y)$ on $C/N \quad$ [property of limits]
b) $L_1(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_1 \vert_{C/N} = L \leq L$ for $C/N \subset A$]+[linearity]
c) $L_1(0,y) = L(0,y) = L(y)$ on $(0,C/N) \subset A \quad$ [first equation in $A$, second in $C/N$]
d) $L_2(0,y) = L(y)$ on $(0,A) \quad$ [Hahn-Banach with $L_2 \vert_{(0,C/N)} = L_1 \leq L$ for $(0,C/N) \subset (0,A)$]+[linearity]
e) $L_3(y) = L(y)$ on $A \quad$ [Hahn-Banach with $L_3 \vert_{(0,A)} = L_2 \leq L$ for $(0,A) \subset A$]+[linearity]

Now $L_3(y) = L(y)$ on $A$ and $L_3(0,y) = L_3 \vert_{(0,A)} (0,y) = L_2 (0,y) = L(y)$.
Since both are equal to $L(y) \;$, $\; L_3(0,y) = L_3(y)$ for all $y \in A$.

But as you write it now, you define $L_1=L$ on $C/N$ and extend with Hahn-Banach so that $L_1=L$ on all of $A$, so there is no point in defining $L_1$ at all. Likewise with the definitions of $L_2$ and $L_3$. All of them are defined as equal to the previous one at some subspace, and then it turns out that it is equal to the previous one at all of $A$. So, $L_3=L_2=L_1=L$, and nothing is gained.

But the real error, this time, is in d). It does not follow from your application of Hahn-Banach there that $L_2(0,y)=L(y)$ on $(0,A)$, it just follows that $L_2(0,y)=L(0,y)$ for all $y\in A$. But the conclusion of d) is essential for the argument, in the line $L_3(0,y) = L_3 \vert_{(0,A)} (0,y) = L_2 (0,y) = L(y)$.

In your previuos version you had $L_2(0,y)=L(y)$ for $y\in A$, but then, you couldn't obtain $L_2(0,y)=L(0,y)$ instead. You simply cannot prove that $L_3(0,y)=L_3(y)$ anywhere outside $C/N$ (and likewise for $L_1$ and $L_2$).

 

[fresh_42]

Thank you. A closer look on how to apply Hahn-Banach finally showed me the trap I've fallen into. It doesn't matter whether there is a difference between my previous or my last version, which I did not intend to be. At some point I trick myself while losing the $0$. It doesn't matter where.
So let me think about one of my other ideas with Hamel basis and continuity, resp. a limit procedure.
Sorry for your inconvenience. At least it showed that the most "obvious" results are often the hardest to prove. (Of course mostly only until one has found the solution.)

 

[Erland]

Ok, we all make mistakes...

My idea is to look at the proof of Hahn-Banach's Theorem and see if it could be modified, for this case, so that 5) holds. No success yet...

 

[fresh_42]

Ok, we all make mistakes...

My idea is to look at the proof of Hahn-Banach's Theorem and see if it could be modified, for this case, so that 5) holds. No success yet...

The problem with my approach is, that $L$ is already fixed. However, $L$ is linear, continuous and behaves well on $\mathbb{R}$ and $||L||_{\infty} = 1$. There must be a way to show it. I mean what more can I hope for?
Another approach could be by starting with the shift invariance, linearity and $||\, .\, ||_{\infty}$ as sublinear upper bound, apply Hahn-Banach and deduce the rest. Perhaps it's easier this way round.

 

[Erland]

Ok, I did a Google search and found a paper by Abraham Robinson, from which it is clear that generalized limits can be constructed by using nonstandard analys (several of them, thus solving also Advanced Problem 7 a, in the September Challenge, and probably b, c, and d can also be solved by this method). This proof would be really simple, if one is aquainted to nonstandard analysis, that is . Unfortunately, I cannot give the proof here, since that would be cheating.

And the axiom of choice shows up indirectly here too, in the construction of a nonstandard extension.

 

[fresh_42]

Ok, I did a Google search and found a paper by Abraham Robinson, from which it is clear that generalized limits can be constructed by using nonstandard analys (several of them, thus solving also Advanced Problem 7 a, in the September Challenge, and probably b, c, and d can also be solved by this method). This proof would be really simple, if one is aquainted to nonstandard analysis, that is . Unfortunately, I cannot give the proof here, since that would be cheating.

And the axiom of choice shows up indirectly here too, in the construction of a nonstandard extension.

I still hope to find an elegant way to prove it after the choice of $L$. Otherwise one would have to start with the shifting invariance and deduce the rest. However, it looks so simple ...

 

[Erland]

I finally have a proof of Advanced Problem 4. But in order to decide whether I have followed the challenge rules or not, I need to tell you how I found it:

I googled "generalized limit" and found an article by Abraham Robinson, where he uses nonstandard analysis to construct a particular class of generalized limits. I didn't read all the details, but I realized from this how one can construct generalized limits using nonstandard analysis in some simple special cases of Robinson's class. This gives a short proof of the existence of generalized limits. But I didn't want a nonstandard proof, so I tried to convert this proof into a "standard" proof. This is possible for all nonstandard proofs, but certainly not easy. The proof given here is the result of this "conversion". It is not strictly the converted version of the nonstandard proof, but rather inspired by it.
Although I probably wouldn't have found this proof if I hadn't found Robinson's article, I still don't think it is against the rules to publish my proof here, since it took me considerable effort to construct the standard proof. Besides, we are halfway into September and the August Challenge should be expired, and I don't care if I get the credit for this solution or not, I just want to share it with you.

The proof is based upon ultrafilters. If you are not aquainted to ultrafilters, you may look up this paper I once wrote myself about nonstandard analysis. Chapter 5 (p. 63 ff.) gives the basic properties of filters and ultrafilters. In particular, look up Definitions 5.1, Corollary 5.7, and Theorem 5.8. These will be used in the proof below without referring to them.
Notice that the axiom of choice is used in Theorem 5.8 (about extending a filter to an ultrafilter), so this axiom is essential here too.

Lemma 1.

Assume that:
1. $I$ is a nonempty set.
2. $\mathcal U$ is an ultrafilter on $I$.
3. $(K,d)$ is a compact metric space.
4. $f: I\to K$ is a function.

Then, there is a unique $x\in K$ such that for every $\epsilon >0$, $f^{-1}(B(x,\epsilon))\in \mathcal U$ (where $B(x,\epsilon)=\{y\in K\,|\,d(y,x)<\epsilon\}$).

Proof: Since $K$ is a compact metric space, it is also totally bounded, which implies that to each integer $n>0$ there is a finite subset $\{x_1^n,x_2^n,\dots,x_{m_n}^n\}\subseteq K$ such that $\cup_{k=1}^{m_n} B(x_k^n,1/n)=K$. This gives $\cup_{k=1}^{m_n}f^{-1}(B(x_k^n,1/n))=I$. Since $\mathcal U$ is an ultrafilter, this means that for some $k$ ($1\le k\le m_n$), $f^{-1}(B(x_k^n,1/n))\in \mathcal U$. Put $x_n=x_k^n$ for every $n>0$, so that we obtain a sequence $\{x_n\}_{n=1}^\infty$ in $K$.
Since $K$ is compact, this sequence has a convergent subsequence $\{x_{n_l}\}_{l=1}^\infty$, with a limit $x\in K$.
Now, take $\epsilon>0$. There is an $l>0$ such that both $n_l\ge 2/\epsilon$ and $d(x_{n_l},x)<\epsilon/2$ hold. Then $B(x_{n_l},1/n_l)\subseteq B(x_{n_l},\epsilon/2)\subseteq B(x,\epsilon)$, which implies $f^{-1}(B(x_{n_l},1/n_l))\subseteq f^{-1}(B(x,\epsilon))$. Since $f^{-1}(B(x_{n_l}))\in\mathcal U$, $f^{-1}(B(x,\epsilon))\in\mathcal U$. This holds for all $\epsilon>0$.
It remains to prove that this $x$ is unique: Assume that $y$ also has these properties, with $y\neq x$. Put $\epsilon=d(x,y)/2>0$. Then also $f^{-1}(B(y,\epsilon))\in\mathcal U$.
Then $\varnothing = f^{-1}(B(x,\epsilon))\cap f^{-1}(B(y,\epsilon))\in\mathcal U$, which is impossible, since $\mathcal U$ is an ultrafilter.
Hence, this $x$ is unique, which completes the proof of Lemma 1.

Next, let $\mathcal V$ be the family of all cofinite subsets of $\mathbb Z_+$, i.e. the family of all $A\subseteq \mathbb Z_+$ such that $\mathbb Z_+\setminus A$ is finite. It is easy to verify that $\mathcal V$ is a filter on $\mathbb Z_+$. Then, there is an ultrafilter $\mathcal U$ on $\mathbb Z_+$ such that $\mathcal V\subseteq \mathcal U$.
We fix such a $\mathcal U$.

$\mathcal U$ contains no finite sets: if $A\in \mathcal U$ is finite, then $\mathbb Z_+\setminus A\in \mathcal V\subseteq \mathcal U$, and hence $\varnothing=A\cap \mathbb Z_+\setminus A\in\mathcal U$, which is a contradiction.

Let $\{x_n\}_{n=1}^\infty$ be a bounded a sequence in $\mathbb R$, say that $|x_n|\le M$ for all $n>0$. The closed interval $[-M,M]$ is a compact metric space (with the metric given by $|x-y|$). Also, the sequence can be considered as a function $f:\mathbb Z_+\to [-M,M]$, with $f(n)=x_n$ for all $n>0$.
Now, it follows from Lemma 1 that there exists a unique $x\in [-M,M]$ such that for all $\epsilon >0$ $\{n\in\mathbb Z_+\,|\,|x_n-x|<\epsilon\}\in\mathcal U$.
This $x$ is independent of the bound $M$: if $M'$ is another bound such that $|x_n|\le M'$ for all $n>0$, with $M\le M'$, say, and the we obtain $y$ instead of $x$ if we apply the above to $M'$ instead of $M$, then also $x\in [-M',M']$, so by the uniqueness for the $M'$ case, $x=y$. Likewise if $M'\le M$.
It follows that there is a unique $x\in \mathbb R$ such that such that for all $\epsilon >0$, $\{n\in\mathbb Z_+\,|\,|x_n-x|<\epsilon\}\in\mathcal U$.

Such a unique $x$ exists for every bounded real sequence $\{x_n\}_{n=1}^\infty$, so this defines a functional $F:\mathcal B\to R$, where $\mathcal B$ is the set of all bounded real sequences $\{x_n\}_{n=1}^\infty$: $F(\{x_n\}_{n=1}^\infty)=x$, with this $x$ just described.

We next prove that $F$ is a (real) linear functional (with the obvious addition and multiplication by scalar): Let $a,b\in \mathbb R$ and $\{x_n\}_{n=1}^\infty,\,\{y_n\}_{n=1}^\infty\in\mathcal B$, with $F(\{x_n\}_{n=1}^\infty)=x$ and $F(\{y_n\}_{n=1}^\infty)=y$. Take $\epsilon>0$.
Put $A=\{n\in\mathbb Z_+\,|\,|x_n-x|<\frac\epsilon{2|a|+1}\}$ and $B=\{n\in\mathbb Z_+\,|\,|y_n-y|<\frac\epsilon{2|b|+1}\}$. Then $A,B\in \mathcal U$, and hence $A\cap B\in \mathcal U$. If $n\in A\cap B$, then $|(ax_n+by_n)-(ax+by)|\le |a||x_n-x|+|b||y_n-y|<|a|\frac\epsilon{2|a|+1}+|b|\frac\epsilon{2|b|+1}<\epsilon$. This holds for all $n\in A\cap B$, so $A\cap B\subseteq \{n\in\mathbb Z_+\,|\,|(ax_n +by_n)-(ax+by)|<\epsilon\}$. Therefore, $\{n\in\mathbb Z_+\,|\,|(ax_n +by_n)-(ax+by)|<\epsilon\}\in \mathcal U$. This holds for all $\epsilon>0$. By the uniqueness clause above and the subsequent definition of $F$, this means that $F(a\{x_n\}_{n=1}^\infty+ b\{x_n\}_{n=1}^\infty)=F(\{ax_n+by_n\}_{n=1}^\infty)=ax+by$.

So, $F$ is linear.

Given $\{x_n\}_{n=1}^\infty\in \mathcal B$, we have $F(\{x_n\}_{n=1}^\infty)\le \overline \lim x_n$: If $x>\overline\lim x_n$, put $\epsilon=|x-\overline \lim x_n|/2$. Then $\{n\in\mathbb Z_+\,|\,|x_n-x|<\epsilon\}$ is a finite set, and then it does not lie in $\mathcal U$, so $F(\{x_n\}_{n=1}^\infty)\neq x$.
Applying this and linearity, we obtain $F(\{x_n\}_{n=1}^\infty)=-F(-\{x_n\}_{n=1}^\infty)=-F(\{-x_n\}_{n=1}^\infty)\ge -\overline \lim (-x_n)=\underline\lim x_n$.
Thus, $\underline \lim x_n \le F(\{x_n\}_{n=1}^\infty)\le \overline\lim x_n$.

Now, we can prove the following:

Lemma 2.

Let $G:\mathcal B\to\mathcal B$ be linear operator such that
1) $\underline\lim x_n\le \underline\lim y_n\le \overline\lim y_n\le\overline\lim x_n$, and
2) $\lim_{n\to\infty}(z_n-y_n)=0$,
where $\{y_n\}_{n=1}^\infty=G(\{x_n\}_{n=1}\infty)$, $\{z_n\}_{n=1}^\infty =G(\{u_n\}_{n=1}^\infty)$, and $u_n=x_{n+1}$ for all $n\in\mathbb Z_+$, for
$\{x_n\}_{n=1}^\infty)\in\mathcal B$.

Then $L=F\circ G:\mathcal B\to \mathbb R$ is a generalized limit.

Proof: Since $F$ and $G$ are linear, so is $L=F\circ G$. With $\{x_n\}_{n=1}^\infty$, $\{y_n\}_{n=1}^\infty$, $\{u_n\}_{n=1}^\infty$, and $\{z_n\}_{n=1}^\infty$ related as above, we have $\underline \lim y_n \le F(\{y_n\}_{n=1}^\infty)\le \overline\lim y_n$. Since $L(\{x_n\}_{n=1}^\infty)=F(\{y_n\}_{n=1}^\infty)$, 1) gives $\underline \lim x_n \le L(\{x_n\}_{n=1}^\infty)\le \overline\lim x_n$.

Now, put $F(\{y_n\}_{n=1}^\infty)=y$. Let $\epsilon>0$. Put $A=\{n\in\mathbb Z_+\,|\,|y_n-y|<\epsilon/2\}$. Then $A\in\mathcal U$. By 2), there is an $N\in\mathbb Z_+$ such that $|z_n-y_n|<\epsilon/2$ for all $n\ge N$. Put $B=\{n\in \mathbb Z_+\,|\,n\ge N\}$. Then $B\in\mathcal U$, since $B$ is cofinite. Now, if $n\in A\cap B$, $|z_n-y|\le|z_n-y_n|+|y_n-y|<\epsilon/2 + \epsilon/2=\epsilon$. This means that $A\cap B\subseteq \{n\in \mathbb Z_+\,|\,|z_n-y|<\epsilon\}$, so $\{n\in \mathbb Z_+\,|\,|z_n-y|<\epsilon\}\in\mathcal U$. This holds for all $\epsilon >0$, which means that $F(\{z_n\}_{n=1}^\infty)=F(\{y_n\}_{n=1}^\infty)$, or $L(\{u_n\}_{n=1}^\infty)= L(\{x_n\}_{n=1}^\infty)$.
This holds for all $\{x_n\}_{n=1}^\infty\in \mathcal B$ and $\{u_n\}_{n=1}^\infty$, such that $u_n=x_{n+1}$ for all $n\in\mathbb Z_+$.

We have now proved that $L$ satisfies conditions 1, 2, 3, and 5 in the definition of a generalized limit.
4 follows from 3, since $\underline \lim x_n\ge 0$ if $x_n\ge 0$ for all $n \in\mathbb Z_+$.
6 also follows from 3, for if $\lim_{n\to \infty} x_n$ exists, then $\lim_{n\to \infty} x_n=\underline \lim x_n\le L(\{x_n\}_{n=1}^\infty)\le\overline\lim x_n = \lim_{n\to \infty} x_n$.
Thus, $L(\{x_n\}_{n=1}^\infty)=\lim_{n\to \infty} x_n$, that is, 6 holds.

Thus $L$ is a generalized limit, so Lemma 2 is proved.

It remains to prove that there exists an operator $G$ which satisfies the conditions in Lemma 2. Indeed, if we could find $G$ and $G'$ both satisfying these conditions, such that $L=F\circ G\neq F\circ G´=L'$, then we have found two different generalized limits, thus solving Advanced Problem 7a in the September Challenge.

I leave the latter problem for the moment, and confine myself to find just one $G$:

Define $G:\mathcal B\to\mathcal B$ by $G(\{x_n\}_{n=1}^\infty)=\{y_n\}_{n=1}^\infty$, for $\{x_n\}_{n=1}^\infty\in\mathcal B$, where $y_n=\frac1n\sum_{k=1}^n x_k$, for all $n\in\mathbb Z_+$.
It is clear that $\sup_n |y_n|\le\sup_n |x_n|$, so $\{y_n\}_{n=1}^\infty\in\mathcal B$.
It is also clear that $G$ is linear. We must prove that $G$ satisfies 1) and 2) of Lemma 2.

Given $\{x_n\}_{n=1}^\infty\in\mathcal B$, let $\{y_n\}_{n=1}^\infty$ be as above and $\{u_n\}_{n=1}^\infty$ and $\{z_n\}_{n=1}^\infty$ as in Lemma 2, related to these$\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$.
Put $s=\overline \lim x_n$.
Take $\epsilon>0$. Put $M=\max(\epsilon/4,\sup_n |x_n|)$. There is an $N_1\in\mathbb Z_+$ such that $x_n<s+\epsilon/4$ for all $n\ge N_1$.
Choose $N\in \mathbb Z_+$ such that $N\ge 4MN_1/\epsilon\ge N_1$.
Then, for $n\ge N$: $y_n=\frac1n\sum_{k=1}^n x_k=\frac1n\sum_{k=1}^{N_1} x_k +\frac1n\sum_{k=N_1+1}^n x_k \le$
$\le MN_1/n +(s+\epsilon/4) (n-N_1)/n\le MN_1/N+(s+\epsilon/4)n/n+MN_1/N+(\epsilon/4)N_1/n<$
$<\epsilon/4+(s+\epsilon/4)+\epsilon/4+\epsilon/4=s+\epsilon$.
Thus, $y_n < s+\epsilon$.

There is such an $N$ to each $\epsilon>0$, which means that $\overline \lim y_n\le s=\overline\lim x_n$.
Applying this to $-\{x_n\}_{n=1}^\infty$ and using linearity, we obtain $\underline\lim y_n=-\overline \lim (-y_n)\ge -\overline \lim (-x_n)=\underline\lim x_n$.

It follows that $\underline \lim x_n\le\underline\lim y_n\le\overline\lim y_n\le\overline\lim x_n$. This holds for all $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ related as above, that is, 1) in Lemma 2 holds.

Next, let $\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$ be as above, and $\{u_n\}_{n=1}^\infty$ and $\{z_n\}_{n=1}^\infty$ as in Lemma 2, related to these$\{x_n\}_{n=1}^\infty$ and $\{y_n\}_{n=1}^\infty$.
Then, for $n\in \mathbb Z_+$, and with $M$ as above, $|z_n-y_n|=|\frac1n\sum_{k=1}^n u_n -\frac1n\sum_{k=1}^n x_n|=\frac1n|\sum_{k=2}^{n+1} x_k -\sum_{k=1}^n x_k|= |x_{n+1}-x_1|/n\le 2M/n\to 0$ as $n\to\infty$.

So, $\lim_{n\to\infty}|z_n-y_n|=0$. This holds for all $\{x_n\}_{n=1}^\infty\in\mathcal B$, that is, 2) in Lemma 2 holds.

We have proved that all the conditions of Lemma 2 are satisfied for our $G$. Thus, by Lemma 2, $L=F\circ G$ is a generalized limit.

Thus, there exists a generalized limit, Q.E.D.

 

[micromass]

Very nice! Here is a proof using Hahn-Banach instead:

Consider the set of all bounded sequences $\ell^\infty$. Let $S((x_n)) = x_{n+1}$. Consider
$$W = \{Sx - x~\vert~x\in \ell^\infty\}$$
We clam that for any $y\in W$ and $c\in \mathbb{R}$ we have
$$\|y + c1\|_\infty \geq |c|$$
Indeed, if $z = Sx - x + c1$, then $z_n = x_{n+1} -x_n +c$ and
$$\|z\|_\infty\geq \frac{1}{N}\sum |z_n| \geq \frac{1}{n}\left|\sum z_n\right| = \left|\frac{1}{N}(x_{N+1} - x_1) + c)\right|$$
This converges to $|c|$ as $N\rightarrow +\infty$ because $|x_{N+1} - x_1|\leq 2\|x\|_\infty$ is bounded.

In particular, it follows that the set of constant sequences $C$ has trivial intersection with $W$. Define on $W\oplus C$
$$\Phi_0(y + c1) = c$$
We have just established that $\Phi_0(z)\leq \|z\|_\infty$ for every $z\in W\oplus C$.

Using the Hahn-Banach theorem, there exists a linear functional $\Phi$ on $\ell^\infty$ extending $\Phi_0$ such that
$$|\Phi(x)|\leq \|x\|_\infty$$
This $\Phi$ satisfies (1) and (2) by construction.

We have
$$\Phi(Sx) - \Phi(x) = \Phi(Sx - x) = \Phi_0(Sx - x) = 0$$
This $(5)$ is satisfied.

For $x\geq 0$, we write $x = \|x\|_\infty y$ where $y_n\in [0,1]$ for all $n$. Then
$$\|1 - y\|_\infty \leq 1$$
Hence
$$1 - \Phi(y) = \Phi(1-y)\leq \|1-y\|_\infty\leq 1$$
Implying
$$\Phi(y)\geq 0$$
And thus
$$\Phi(x) = \|x\|_\infty\Phi(y) \geq 0$$
Thus (4) is satisfied.

Given $x\in \ell^\infty$ and any $\alpha,\beta$ with
$$\alpha<\liminf x_n,~\limsup x_n<\beta$$
there is an $N$ such that $\alpha<x_n<\beta$ for all $n>N$. This means that
$$y = S^N x - \alpha 1,~z = \beta.1 - S^Nx$$
are positive bounded sequences. Hence
$$0\leq \Phi(y) = \Phi(S^N x) - \alpha = \Phi(x) -\alpha$$
Implying $\alpha<\Phi(x)$, and
$$0\leq \Phi(z) = \beta - \Phi(S^n x) = \beta - \Phi(x)$$
Implying $\Phi(x)<\beta$.
Since $\alpha,\beta$ are chosen arbitrarily, property (3) follows. Property (6) is a direct consequence from (3).

 

[micromass]

Erland, are you acquainted with convergence of (ultra)filters? I think it would make your proof a lot more cleaner if you could use that instead.

 

[Erland]

Erland, are you acquainted with convergence of (ultra)filters? I think it would make your proof a lot more cleaner if you could use that instead.

I did read about it many years ago, so I had to look it up. Perhaps that could work, but how? Recall that my ultrafilter is not on $\mathbb R$, but on $\mathbb Z_+$, which has no obviuos suitable topology.

Btw, nice Hahn-Banach proof, much simpler than mine!